Problem: Let $f(x, y, z) = \sqrt{xyz}$ Suppose $\vec{a} = (-1, -1, 4)$ and $\vec{v} = \left( \dfrac{\sqrt{2}}{2}, \dfrac{1}{2}, \dfrac{1}{2} \right)$. Find the directional derivative of $f(x, y, z)$ at $\vec{a}$ in the direction of $\vec{v}$. $\dfrac{\partial f}{\partial v} = $
Explanation: The directional derivative of a function $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\left( \nabla f (\vec{a}) \right) \cdot \vec{v}$. Let's find the gradient of $f$. We can use the chain rule, treating whichever two variables we aren't differentiating with respect to as constants. $\begin{aligned} \nabla f &= \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z} \right) \\ \\ &= \left( \dfrac{1}{2} yz (xyz)^{-\frac{1}{2}}, \dfrac{1}{2} xz (xyz)^{-\frac{1}{2}}, \dfrac{1}{2} xy (xyz)^{-\frac{1}{2}} \right) \end{aligned}$ Plugging in $(-1, -1, 4)$, we can find $\nabla f$ at $\vec{a}$. $\begin{aligned} \nabla f (\vec{a}) &= \left( \dfrac{1}{2} (-4) (4)^{\frac{-1}{2}}, \dfrac{1}{2} (-4) (4)^{\frac{-1}{2}}, \dfrac{1}{2} (1) (4)^{\frac{-1}{2}} \right) \\ \\ &= \left( -1, -1, \dfrac{1}{4} \right) \end{aligned}$ Therefore: $\begin{aligned} \left( \nabla f (\vec{a}) \right) \cdot \vec{v} &= \left( -1, -1, \dfrac{1}{4} \right) \cdot \left( \dfrac{\sqrt{2}}{2}, \dfrac{1}{2}, \dfrac{1}{2} \right) \\ \\ &= \dfrac{-\sqrt{2}}{2} - \dfrac{1}{2} + \dfrac{1}{8} \\ \\ &= \dfrac{-\sqrt{2}}{2} - \dfrac{3}{8} \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\dfrac{-\sqrt{2}}{2} - \dfrac{3}{8}$.